Consider filling a cylinder of compressed argon from a high- pressure supply line as shown below. Before filling, the cylinder contains 10 bar of argon at room temperature. The valve is then opened, exposing the tank to a 50 bar line at room temperature un- til the pressure of the cylinder reaches 50 bar. The valve is then closed. For argon take cP = (5/2)R and the molecular weight to be 40 g mol. You may use the ideal gas model. (a) What is the temperature right after the valve is closed? (b) If the cylinder sits in storage for a long time, how much heat is transferred (in kJ/kg)? (c) What is the pressure of the cylinder when it is shipped (after it was stored for a long time)?

The initial pressure in the cylinder of 10 bar of argon with [tex]C_p[/tex] = (5/2)·R

and the added gas at pressure of 50 bar gives the following values.

(a) 400.6 K

(b) 51.184 J

(c) The pressure of the cylinder when it is shipped is 37.194 bar

(a) Adiabatic compression, we have;

[tex]\dfrac{p_{1}}{p_{2}} = \mathbf{\left (\dfrac{T_{_{1}}}{T_{2}} \right )^{\dfrac{k}{k-1}}}[/tex]

Where;

[tex]k = \dfrac{C_p}{C_v} = \mathbf{\dfrac{C_p}{C_p - R}}[/tex]

Therefore;

[tex]K= \mathbf{\dfrac{\frac{5}{2} \cdot R}{\frac{5}{2} \cdot R - R} }= \dfrac{5}{3}[/tex]

Which gives;

[tex]\dfrac{10}{50} = \left (\dfrac{298}{T_{2}} \right )^{\dfrac{\frac{5}{3} }{\frac{5}{3} -1}} = \left (\dfrac{298}{T_{2}} \right )^{2.5}[/tex]

[tex]\dfrac{50}{10} = \mathbf{ \left (\dfrac{T_2}{298.15} \right )^{2.5}}[/tex]

[tex]ln(5)= 2.5\cdot ln\left (\dfrac{T_2}{298.15} \right )[/tex]

[tex]T_2 = 298 \times e^{\dfrac{ln(5)}{2.5} } \approx \mathbf{567.29\, K}[/tex]

The temperature

Using an ideal gas model, we have;

[tex]\dfrac{V_{1}}{V_{2}} = \mathbf{\dfrac{P_{2}\times T_{1}}{T_{2} \times P_1}}[/tex]

Which gives;

[tex]\dfrac{V_{1}}{V_{2}} = \dfrac{50\times 298}{567.29 \times 10} \approx 2.625[/tex]

V₁ = 2.625·V₂

Volume occupied by the gas in the cylinder after the pressure increases is therefore;

[tex]V_2 = \dfrac{V_1}{2.625} \approx 0.381 \cdot V_1[/tex]

Volume of gas added is therefore;

[tex]V =V_1 - \dfrac{V_1}{2.625} = 0.619\cdot V_1[/tex]

Considering a molar volume of gas, we have;

0.040 × 0.619 × 2.5×8.3145×(T₃ - 298) = 0.040 × 0.381 × 2.5×8.3145×(567.29 - T₃)

Solving gives;

T₃ ≈ 400.6 K

For a molar volume of gas cylinder, the temperature just after the valve is closed is T₃ ≈ 400.6 K

(b) If the cylinder sits in the storage for a long time, and heat is conducted out, we have;

The temperature of the cylinder will revert to room temperature of 298 K

The heat transferred is therefore;

0.040 × (2.5 × 8.3145 - 8.3145) ×(400.6 - 298) ≈ 51.184

The heat transferred from the cylinder is 51.184 J

(c) The pressure of the gas in the cylinder is therefore;

[tex]\dfrac{P_1}{T_1} = \mathbf{\dfrac{P_2}{T_2}}[/tex]

Which gives;

[tex]P_2= \mathbf{\dfrac{P_1}{T_1} \times T_2}[/tex]

Therefore;

[tex]P_2= \dfrac{50}{400.6} \times 298 \approx 37.194[/tex]

The pressure of the cylinder when it is shipped is 37.194 bar

Learn more about thermodynamics here:

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